Question: Solve the exponential equation for $x$. ( 1 4 ) 9 x − 5 = 32 x + 8 \left(\dfrac{1}{4}\right)\^{9x-5}=32\^{ x+8} $x=$
The strategy We want to rewrite both of the exponential terms in the equation so that the bases of the two terms are the same. Then, we will be able to equate the exponents and solve for $x$. Matching the bases Let's rewrite ( 1 4 ) 9 x − 5 \left(\dfrac{1}{4}\right)\^{ 9x-5} and 32 x + 8 32\^{ x+8} so their common base is $2$. ( 1 4 ) 9 x − 5 ( 2 − 2 ) 9 x − 5 2 − 18 x + 10 = 32 x + 8 = ( 2 5 ) x + 8 = 2 5 x + 40 ( since 1 4 = 2 − 2 and 32 = 2 5 ) ( ( a n ) m = a n ⋅ m ) \begin{aligned}\left(\dfrac{1}{4}\right)\^{ 9x-5}&=32\^{ x+8}\\\\ (2^{-2})\^{ 9x-5} &=(2^{5})\^{ x+8}&&&&(\text{since }\dfrac{1}{4}=2^{-2} \text{ and } 32= 2^{5}) \\\\ 2\^{-18x+10} &= 2\^{ 5x+40} &&&&((a^n)^m=a^{n\cdot m})\end{aligned} Solving the linear equation We obtain the following equation. 2 − 18 x + 10 = 2 5 x + 40 2\^{-18x+10} = 2\^{ 5x+40} Now we can equate the exponents and solve for $x$. $\begin{aligned} -18x+10&=5x+40\\\\ x &= -\dfrac{30}{23}\end{aligned}$ The answer The answer is $x=-\dfrac{30}{23}$. You can check this answer by substituting $\it{x=-\dfrac{30}{23}}$ in the original equation and evaluating both sides.